# Calculating the density of a joint distribution

For the joint density function $$P\big((X,Y) \in A\big) = \int_A f_{(X,Y)} (x,y) \, dx\,dy$$ how would you show that if $(X,Y)$ is a random vector in $\mathbb{R}^2$ with density $f_{(X,Y)}$ and $f_{(X,Y)}(x,y) = f(x)g(y)$ for a pair of non-negative functions $f$ and $g$ then $X$ has density $$\frac{f}{\int_\mathbb{R}f(t)\,dt}$$ and $Y$ has density $$\frac{f}{\int_\mathbb{R}g(t)\,dt}\, \, ?$$

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If I understand correctly, X and Y would be independent, f and g would be their marginal distributions and would thus be their density functions as well. Integrated over R, f and g would yield an answer of unity and the statement would hold but this doesn't seem right for some reason. Maybe someone will point out why. – Inquest Oct 17 '12 at 19:24
The only thing that, it is not said that $f$ and $g$ are already density functions, so they may differ in a constant $c=\int_{\Bbb R}f$.. – Berci Oct 17 '12 at 19:30

$$\Pr(X\in B) = \Pr((X,Y)\in B\times\mathbb R) = \int\limits_{B\times\mathbb R} f(x)g(y)\,dx\,dy.$$ Do we believe that this double integral is equal to the following iterated integral? $$\int_B \left( \int_{\mathbb R} f(x)g(y)\,dy \right)\,dx$$ Tonelli's theorem will tell you that you have equality here, since $f,g\ge 0$. Now $$\int_{\mathbb R} f(x)g(y)\,dy = f(x)\int_{\mathbb R} g(y)\,dy$$ because $f(x)$ does not depend on $y$. "Constant" always means not depending on something, and there are times when one should explicitly say what something does not depend on. Now we're looking at $$\int_B\left( f(x)\int_{\mathbb R} g(y)\,dy\right)\,dx.$$ . . . and we say: $\int_{\mathbb R} g(y)\,dy$ does not depend on $x$. Therefore it is a "constant" and can be pulled out of $\int\cdots\cdots\,dx$. But I'm not going to do that this time! Instead, let's observe that we have proved $$\Pr(X\in B) = \int_B(\text{a certain function of }x)\,dx.$$ Therefore that certain function of $x$ is the density that we seek.

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Given $f_{(X,Y)}(x,y) = f(x)g(y)$ it looks as though $X$ and $Y$ are independent, and so the density of $X$ is just $f(x)$.

As $\int_{X \le a, y \in \mathbb{R}} f_{(X,Y)}(x,y) = \int_{0}^{a}f(x)\int_\mathbb{R}g(y)dy \ dx$

$= \int_{0}^{a}f(x)dx = F(a)$

So $x$ has density $\frac{d}{dx}(F(x))=f(x)$

Similarly the integrals in the denominators would be unity. Is the numerator in the latter equation intended to be $g$?

Could be wrong though - hope I understood the question:)

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But $f_{(X,Y)}(x,y)$ is also $2f(x)\frac12g(y)$... Hence the density of $X$ is $2f(x)$ as well? – Did Oct 17 '12 at 21:14